0
$\begingroup$

I'm trying to prove this result on my first course on Real Analysis. The proof from totally bounded to bounded is alright, my problem is on the other implication.

$\textbf{Definition of totally bounded set}$: $A$ is totally bounded if there is some fixed $r>0$ and an index $J \in \mathbb{N}$ such that $A \subseteq \cup_{j=1}^J B_{r}(x_{j})$, where $ B_{r}(x_{j})$ is an open ball of radius $r$ centered at $x_{j}$.

First, let $A \subseteq \mathbb{R}^n$ be a bounded set. Then, for any two elements $x,y \in A$, we have that $||x-y|| < C$ for some large enough $C$. Well, if $A$ is finite this is trivial. So, suppose $A$ is infinite. Then, by Bolzano-Weierstrass Theorem, there must be some accumulation point $x_{1}$ of $A$. So, take $B_{r}(x_{1})$. If $A \subseteq B_{r}(x_{1})$, we are done. If not, then there must be some $x_{2} \in A$ such that $||x_{1} - x_{2}|| \geq r$. So, take $B_{r}(x_{2})$. Now, check if $A \subseteq (B_{r}(x_{1}) \cup B_{r}(x_{2}))$. If so, we are done. If not. there must be some $x_{3} \in A$ such that $||x_{3} - x_{2}|| \geq r$ and $||x_{3} - x_{1}|| \geq r$. Now, take $B_{r}(x_{3})$ and so on...

My intuitive argument is that you have to stop with a finite number of balls. We are covering a bigger and bigger part of set $A$. Since it's limited, the required distance between $x_{1}$ and the next candidate to be the center of a new ball will eventually be bigger than $C$, which can't happen. Hence, the above procedure must end with a finite number of balls. Otherwise, we could fill a given box with and infinite number of tenis balls of radius $r>0$ LOL!

Any ideas how to go on with this proof? I'm not supposed to work with sequences yet, since the sequence chapter of my course is the next one. I can use Heine-Borel and Bolzano-Weierstrass, though. Any ideas?

  • 0
    If this process does not stop, then you have a bounded sequence $(x_n)$ with no convergent subsequence. This contradicts the Bolzano-Weierstrass theorem. – 2017-02-02
  • 0
    Wow, nice hint, @SangchulLee! This sequence would not converge because the elements of any subsequence are separated from each other by a distance of at least $r>0$? – 2017-02-02
  • 0
    Yes, exactly. :) – 2017-02-02
  • 1
    Thanks a lot, sir! May the Force be with you @SangchulLee! – 2017-02-02

1 Answers 1

0

One easy proof that every bounded set $A\subset \Bbb R^n$ is totally bounded is:

$a\mbox{ bounded } \implies \exists C \in \Bbb R: \forall (a\in A,b\in A) ||a-b|| < C; \mbox{ choose }C_0 \mbox{ to be such a }C. $

$ \mbox{ Choose } p \in A \mbox{ and let } J=1,x_1 = p, r=2C_0. $

Then for all $a \in A$ $$ ||a-p|| = ||x_1 - p|| < C_0 < r $$